Pi and billiard balls; a different application of π for Pi Day

      So suppose there is some sort of physical experiment, and the first time you do it, the answer comes out as 3. Then you change the experiment to make it a little more complicated, and the answer comes out as 31. Then you make it even more complicated, and the answer comes out 314. Then you ... . Of course you see where I'm going with this.

      It's March 14, or 3 14 (in the US date format), and because 314 are the first digits of π, many people use this day to share interesting and unusual appearances of π. I think you will find Pi and the Billiard Balls to be a little different!

      But first, if your statistics course did not require calculus as a prerequisite, you may be unaware that π is contained within the formula for the probability density of the Normal distribution, f(x) = $\frac{1}{\sqrt{2\pi {\sigma }^2}}{}^{\exp }$ where the 2π is necessary for the integral of the pdf to equal 1.

      A common application of π on Pi Day is the Buffon's needle problem: Given a needle of length l dropped on a plane ruled with parallel lines t units apart, what is the probability that the needle will lie across a line upon landing? There are many references to this such as Buffon, so the solution will not be repeated here.

      π appears in many places in math. One reason is that π is defined in reference to a circle. The trigonometric functions can be defined in terms of triangles within a circle. As a point traverses the circumference of the circle more than once, the trigonometric functions repeat in cycles of 2𝜋. Therefore, phenomena that repeat and can be represented by trigonometric functions are likely to have π somewhere within them. This is because π inherently relates to the periodic nature of these functions, making it indispensable in modeling cyclical behaviors.

      I recently discovered a physics problem called Pi and Billiard Balls. The original article is by G. Galperin, and I will try to summarize it. His paper is not an easy read.

      Suppose we have the first quadrant of an xy coordinate system. Suppose we have a vertical wall at x = 0, y ⪰ 0. We have ball 1 having mass m1 at initial position x1 > 0, and ball 2 having mass m2 ⪰ m1 at initial position x2 > x1. Let the ratio of the masses be a multiple of 100:   m2 / m1 = 100^N for a fixed non-negative integer N, including N = 0. Suppose ball 2 moves from right to left along the x-axis and collides with ball 1, ball 1 moves from right to left and collides with the vertical wall, and assume all collisions will be perfectly elastic. This perfect elasticity assumption implies the balls will satisfy the law of conservation of momentum, and the law of conservation of kinetic energy. We also assume the balls only move along the x-axis; there is no y movement.

      The amazing conclusion of all this is: The total number of collisions C(N) is a number equal to the first N decimal digits of the number π (starting with 3) !

      For case 1, let N = 0 so m2 = m1. Now push ball 2 from right to left at initial velocity v2 until it hits ball 1. This is collision 1. Ball 1 will move from right to left at the same velocity v1' = v2, while ball 2 will now be at rest. Eventually ball 1 hits the wall for collision 2 and now moves from left to right at velocity -v1', until it hits ball 2 for collision 3. Now ball 1 is at rest, and ball 2 is moving to the right with velocity −v1'. There have been 3 collisions (the first digit of π), and there will be no more.

      For case 2, let N = 1 so m2 / m1 = 100. This is more complicated because after collision 1, v1' will not equal v2, and ball 2 will not be at rest. The result of the two conservation equations m1v1 + m2v2 = m1v1' + m2v2' and .5m1(v1^2) + .5m2(v2^2) = .5m1(v1' ^2) + .5m2(v2' ^2), give v1' = [(m1 - m2)v1 + 2m2v2] / (m1 + m2) and v2' = [(m2 - m1)v2 + 2m1v1] / (m1 + m2). After collision 1, substituting v1 = 0 and m2 / m1 = 100 gives v1' = 200v2/101 ≈ 1.98v2 (nearly twice as fast as initial velocity v2) and v2' = 99v2/101 ≈ .98v2 (slightly less than initial velocity v2).

      Ball 1 hits the wall for collision 2 and now moves from left to right at velocity -v1', until it hits ball 2 for collision 3.

      After collision 3, substituting primes into the two conservation equations, gives v1''' ≈ -0.96v2 and v2''' ≈ 1.96v2. Ball 1 will bounce back and forth between ball 2 and the wall many times. After each collision between the two balls, the velocity of each ball changes with ball 1 decreasing in velocity and ball 2 increasing in velocity. The relative velocity between ball 1 and ball 2 decreases with each collision, as the heavier ball will slowly transfer its momentum to the lighter ball. Eventually the relative velocity will be so small that they will effectively move together after the collision. At this point, no more collisions will occur. There will be 31 collisions (the first two digits of π).

      For case 3, let N = 2 and m2 / m1 = 100², there are 314 collisions. And so on.

      Not surprisingly there is a circle lurking under all of this, due to the conservation of kinetic energy equation .5m1(v1^2) + .5m2(v2^2) = constant, and there is a trigonometric function and a calculation involving an angle and π. Galperin proves the conclusion in general: The total number of collisions C(N) is a number equal to the first N decimal digits of the number π (starting with 3) !

      In addition to the Galperin paper, a good explanation is here, but this is not an easy read.

      You are welcome to try this experiment yourself, if you can create the condition of perfect elasticity. Happy Pi Day.